How Does Substituting t=∞ Affect the Fourier Transform of g(t)?

[frenzal_dude]

Homework Statement

Hi, I need to find the Fourier Transform of:

g(t) = (e^-t)Sin(Wct)u(t)

where Wc=2πFc
and u(t) is the step function which is equal to 1 if time is +ve and 0 otherwise.

Homework Equations

I know that g(t) = (e^-t)[e^jWct - e^-jWct]/2j = [e^-t(1+jWc) - e^t(-1-jWc)]/(2j)
(0<=t<=∞, because of step function u(t))

The Attempt at a Solution

Therefore the Fourier Transform would be:
[1/(2j)]*∫([e^-t(1+jWc) - e^t(-1-jWc)])(e^-jWt)dt

= [1/(2j)]*∫([e^-t(1-jWc+jW) - e^t(-1-jWc-jW)])dt (limits: t=∞ to t=0)

= [1/(2j)][(e^-t(1-jWc+jW))/(-(1-jWc+jW)) - (e^t(-1-jWc-jW))/(-1-jWc-jW)] (sub in: t=∞ to t=0)

If you sub t=+/-∞, the exponential could be 0 or it could be infinite depending on whether 1-jWc+jW and -1-jWc-jW are -ve or +ve.
How can we know if they are positive or negative?

Hope you guys can help!
David.

 

[gabbagabbahey]

I'd assume that Wc ($\omega_c$?) is real-valued.

[tex]\lim_{t\to\infty}e^{-zt}=0[/itex]

So long as the real part of $z$ is positive.

Also, if $g(t)$ is zero for negative $t$, why is one of your integration limits $-\infty$?

 

[frenzal_dude]

I'd assume that Wc ($\omega_c$?) is real-valued.

[tex]\lim_{t\to\infty}e^{-zt}=0[/itex]

So long as the real part of $z$ is positive.

Also, if $g(t)$ is zero for negative $t$, why is one of your integration limits $-\infty$?

so is it also true to say: [tex]\lim_{t\to\infty}e^{zt}=0[/itex] so long as the real part of $z$ is negative?

If that's true that helps me out A LOT! This is something not even my Signal Processing lecturer could explain to me, he told me 'he'd get back to me' and he never did.

Thankyou so much for your help!

Btw I fixed up the limits, you're right it's from t=infinity to t=0

 

[frenzal_dude]

btw this is the answer I got:
$$\frac{\omega + j}{(\omega_c-w)^2+1}$$

 

[gabbagabbahey]

so is it also true to say: [tex]\lim_{t\to\infty}e^{zt}=0[/itex] so long as the real part of $z$ is negative?

Yes, it's fairly easy to show simply by splitting $z$ into real and imaginary parts. Say $z=x+jy$, where $x$ and $y$ are real-valued. Euler's formula then tells you$$e^{zt}=e^{(x+jy)t}=e^{jyt}e^{xt}=\cos(yt)e^{xt}+j\sin(yt)e^{xt}$$

As $t\to\infty$, both $\sin(yt)$ and $\cos(yt)$ oscillate between $0$ and $1$ more and more rapidly, and hence are bounded. Meanwhile, $e^{xt}$ goes rapidly to zero if $x<0$ and so its product with any bounded function will also approach zero.

btw this is the answer I got:
$$\frac{\omega + j}{(\omega_c-w)^2+1}$$

That doesn't look quite right, you'd better show your calculations.

 

[frenzal_dude]

Here's my working out:

$$g(t)=e^{-t}sin(\omega _ct)u(t)= e^{-t}(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct})u(t)$$

 

[frenzal_dude]

$$\therefore G(f) =\frac{1}{2j}\int_{0}^{\infty }(e^{j\omega _ct-t}-e^{-j\omega _ct-t})e^{-j\omega t}dt=\frac{1}{2j}\int_{0}^{\infty }e^{j\omega_ct-t-j\omega t}-e^{-j\omega_ct-t-j\omega t}dt$$

 

[frenzal_dude]

$$\frac{1}{2j}\int_{0}^{\infty }e^{j\omega_ct-t-j\omega t}-e^{-j\omega_ct-t-j\omega t}dt=\frac{1}{2j}[\frac{e^{t(j(\omega_c-w)-1)}}{j\omega_c-1-j\omega}-\frac{e^{-t(j(\omega_c+w)+1)}}{-j\omega_c-1-j\omega}]t=\infty ,t=0$$

 

[frenzal_dude]

$$=-\frac{1}{2j}[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}]=\frac{\omega_c+\omega+j}{2(\omega_c+\omega)^2+2}+\frac{\omega_c-\omega-j}{2(\omega_c-\omega)^2+2}$$

 

[gabbagabbahey]

Here's my working out:

$$g(t)=e^{-t}sin(\omega _ct)u(t)= e^{-t}(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct})$$

Ermm... you mean $g(t)=e^{-t}\left(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct}\right)u(t)$, right? You can't get rid of the $u(t)$ until you put it into the integral and use the fact that $\int_{-\infty}^{\infty}f(t)u(t)dt=\int_{0}^{\infty}f(t)dt$

$$=-\frac{1}{2j}[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}]=\frac{\omega_c+\omega+j}{2(\omega_c+\omega)^2+2}+\frac{\omega_c-\omega-j}{2(\omega_c-\omega)^2+2}$$

This looks fine, but it doesn't simplify to what you wrote originally.

 

[frenzal_dude]

do you mean what I wrote originally by this: $$\frac{\omega + j}{(\omega_c-w)^2+1}$$

I realized where I went wrong and tried it again.

With my final answer, which I hope is now correct, I tried getting a common denominator to make it into one fraction but the fraction ended up being even more complicated, so I thought I'd just leave the answer as how I gave it before

(btw here was the fraction I got when I simplified it: $$\frac{\omega_c(\omega_c^2+\omega^2+1)-2\omega^2-2\omega j}{(\omega_c^2+\omega^2+1)-4\omega^2}$$)

 

[gabbagabbahey]

Your answer from post#9 is correct.

Personally, I wouldn't worry so much about making the denominator real and instead concentrate on putting it over a common denominator:

$$\begin{aligned}-\frac{1}{2j}\left[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}\right] &= -\frac{1}{2j}\left[\frac{j(\omega_c+\omega)+1+j(\omega_c-\omega)-1}{\left(j(\omega_c-\omega)-1\right)\left(j(\omega_c+\omega)+1\right)}\right] \\ &= -\frac{1}{2j}\left[\frac{2j\omega_c}{j^2(\omega_c^2-\omega^2)-2j\omega-1}\right] \\ &= \frac{\omega_c}{\omega_c^2-\omega^2+2j\omega+1} \end{aligned}$$

 

[frenzal_dude]

Thanks for the help.

Quick question in regards to $$\lim_{t\to\infty}e^{-zt}=0$$

What if z is only imaginary? ie. the real part is 0.

How would you sub t=∞ into this: $$e^{j(\omega_1 t+\Theta_1)-jwt}$$

 

[gabbagabbahey]

If the real part is zero, then the limit won't exist. You can use Euler's formula to show this.